Arrays Questions and Answers

Answer: C

Correct answer: Both A and B

A is true: an element of an array may itself be an array, which is exactly what a two-dimensional array is. B is true in the same sense - C has only one-dimensional arrays, and int a[3][4] is an array of 3 objects each of type int[4]. The remaining option is false: array elements always occupy contiguous memory, which is what makes index arithmetic possible.

Answer: C

Correct answer: int arr[] = new int[3];

In Java an array object is created with new followed by the type and the length in square brackets: int arr[] = new int[3];. Round brackets after a type mean a constructor call, and int is a primitive with no constructor. int arr[]; alone only declares the reference - no array exists until new runs.

Answer: B

Correct answer: False

\(O(1)\) means the work does not grow with the input at all, while \(O(N)\) grows in proportion to it. For any large enough \(N\) the constant-time operation wins, so the statement is false. This is why a hash lookup is preferred to a linear scan.

Answer: B

Correct answer: Run-time

The exception occurs at run time. The compiler generally cannot know what an index will evaluate to - it may come from a variable, from input, or from a computation - so the check is made by the JVM as each access executes, and the failure surfaces only when that line runs.

Answer: D

Correct answer: All of above

All three will do. A two-dimensional array of characters gives every name the same fixed width. A one-dimensional array of strings is the same structure described differently. An array of pointers to char lets each name occupy only as much space as it needs, and makes sorting cheap, since only pointers are swapped.

Answer: B

Correct answer: &A[0][0][0] + w(y * z * p + z*q + r)

In row-major order the last index varies fastest. Moving one step in \(p\) skips a whole \(y \times z\) plane, one step in \(q\) skips a row of \(z\) elements, and one step in \(r\) skips a single element:

where \(w\) is the size of one element. The dimension \(x\) never appears, which rules out two of the options immediately - the number of planes does not affect where any one of them starts.

Answer: D

Correct answer: C's shortcoming

C performs no bounds check, so an out-of-range subscript silently reads or writes memory that belongs to something else. It is generally counted a shortcoming: it is the root of buffer overflows and of a great many hard-to-find bugs. The design was deliberate - omitting the check keeps array access as fast as raw address arithmetic - but the cost in safety is the reason later languages put the check back.

Answer: B

Correct answer: No

They hold the same address but are not the same expression: arr has type int * while &arr has type int (*)[10]. The difference appears in arithmetic - arr+1 advances by one integer, &arr+1 by the whole array of ten.

Answer: A

Correct answer: Compiler Error

An array is not assignable. b = a; fails because b is not a modifiable lvalue - the name denotes fixed storage, not a pointer variable that could be repointed. Copying an array means copying its elements, with a loop or with memcpy(b, a, sizeof(a)).

Answer: B

Correct answer: ptr is a pointer to an array of 10 integers

Read the declaration from the identifier outwards, respecting the parentheses. (*ptr) says ptr is a pointer; [10] then says to an array of 10; int gives the element type. Without the parentheses, int *ptr[10] would bind [10] first and mean an array of 10 pointers - the two are quite different, and ptr + 1 advances by 40 bytes in the first case and by one pointer in the second.